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$f(z)$ is analytic if $u_x = v_y$ and $u_y = -v_x$ where $u(z)$ and $v(z)$ are the real and complex part, respectively, of $f(z)$. However, $\frac{1}{z^3+z} = \frac{1}{(x+iy)^3+x+iy}$ doesn't seem to have a simple composition of a real and a complex part. Using polar form doesn't seem to make it easier either. Is there perhaps another way of showing it? It seems it should be an easier way, e.g. if $g(z) = z^3+z$ and $h(z) = 1/z$ are analytic, is their composition, $h(g(z))$, necessarily also analytic?

Frank Vel
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    Yes, compositions of holomorphic functions are again holomorphic, and for the derivative you have the chain rule like in real analysis. The proof is also almost identical to the proof in the real case. – Daniel Fischer Oct 03 '16 at 12:22
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    What about the pole $z=0$? – Piquito Oct 03 '16 at 12:37

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